Problem 4
Question
We will see later that the multiplicative group of nonzero elements of a finite field is cyclic. Illustrate this by finding a genentor for this group for the given finite field. $$ \text { Using Fermat's theorem, find the remainder of } 3^{47} \text { when it is divided by } 23 . $$
Step-by-Step Solution
Verified Answer
The remainder is 4.
1Step 1: Identify Fermat's Little Theorem
Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \pmod{p} \). Here, \( a = 3 \) and \( p = 23 \). Thus, \( 3^{22} \equiv 1 \pmod{23} \).
2Step 2: Simplification Using Exponentiation
To find \( 3^{47} \mod 23 \), first express 47 in terms of multiples of 22: \( 47 = 2\times22 + 3 \). Therefore, \[ 3^{47} = (3^{22})^2 \times 3^3 \]. Using Fermat's theorem, \( 3^{22} \equiv 1 \pmod{23} \), thus \( (3^{22})^2 \equiv 1 \pmod{23} \). So, \( 3^{47} \equiv 3^3 \pmod{23} \).
3Step 3: Calculate the Remainder
Calculate \( 3^3 \): \( 3^3 = 27 \). Find \( 27 \mod 23 \): divide 27 by 23, giving a quotient of 1 and a remainder of 4. Thus, \( 3^{47} \equiv 3^3 \equiv 4 \pmod{23} \).
Key Concepts
Finite FieldsCyclic GroupsModular Arithmetic
Finite Fields
A finite field, often referred to as Galois Field, is a mathematical structure with a finite number of elements. These fields are significant in areas like coding theory and cryptography. Here's what makes a finite field interesting:
- In a finite field, there are a limited number of elements. This count is always a power of a prime number, expressed as \( p^n \), where \( p \) is a prime and \( n \) is a positive integer.
- The finite field of order \( p \) is generally denoted as \( GF(p) \) or simply \( \mathbb{F}_p \).
- Addition, subtraction, multiplication, and division (except by zero) of elements in this field always result in another element within the field.
Cyclic Groups
Cyclic groups are fascinating mathematical structures where you can express every element as a power of a single element called a "generator." If an element \( g \) is a generator for a group \( G \), then every element \( h \) in \( G \) can be represented as \( g^k \) for some integer \( k \).
Key features of cyclic groups include:
Key features of cyclic groups include:
- All cyclic groups are inherently abelian, meaning the order of operations doesn't change the result—i.e., \( a \cdot b = b \cdot a \).
- The structure of cyclic groups is determined by the number of elements they contain. In finite cyclic groups, beyond a certain point, operations wrap around due to modular arithmetic.
- In the context of finite fields, the non-zero elements naturally organize into a cyclic group, often demonstrated using Fermat's Little Theorem.
Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. It's akin to the everyday experience of a clock, where you reset to earlier numbers after reaching 12.
In modular arithmetic:
In modular arithmetic:
- An expression like \( a \equiv b \pmod{n} \) implies that when you divide \( a - b \) by \( n \), you get an integer with no remainder.
- This system solves many problems efficiently, as noticed in the solution of \( 3^{47} \pmod{23} \) by applying Fermat’s Little Theorem to simplify calculations.
- Modular arithmetic plays a crucial role in cryptography, computer science, and other scientific fields requiring repetitive cycles or patterns.
Other exercises in this chapter
Problem 2
We will see later that the multiplicative group of nonzero elements of a finite field is cyclic. Illustrate this by finding a genentor for this group for the gi
View solution Problem 3
We will see later that the multiplicative group of nonzero elements of a finite field is cyclic. Illustrate this by finding a genentor for this group for the gi
View solution Problem 5
We will see later that the multiplicative group of nonzero elements of a finite field is cyclic. Illustrate this by finding a genentor for this group for the gi
View solution Problem 11
$$ \text { In Exercises } 11 \text { through } 18 \text {, describe all solutions of the given congruence, as we did in Examples } 20.14 \text { and 20.15. } $$
View solution